The equation of hyperbola $H$ is $\dfrac {(x-3)^{2}}{49}-\dfrac {(y+6)^{2}}{25} = 1$. What are the asymptotes?
Solution: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y+6)^{2}}{25} = - 1 + \dfrac {(x-3)^{2}}{49}$ Multiply both sides of the equation by $25$ $(y+6)^{2} = { - 25 + \dfrac{ (x-3)^{2} \cdot 25 }{49}}$ Take the square root of both sides. $\sqrt{(y+6)^{2}} = \pm \sqrt { - 25 + \dfrac{ (x-3)^{2} \cdot 25 }{49}}$ $ y + 6 = \pm \sqrt { - 25 + \dfrac{ (x-3)^{2} \cdot 25 }{49}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y + 6 \approx \pm \sqrt {\dfrac{ (x-3)^{2} \cdot 25 }{49}}$ $y + 6 \approx \pm \left(\dfrac{5 \cdot (x - 3)}{7}\right)$ Subtract $6$ from both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{5}{7}(x - 3) -6$